OK, ignore that. I was using i as a loop variable (using i as a Symbol is a
bad idea because of this, imho).
But now I get this strange error:
In [48]: M(i, j, n).args[3]
Out[48]: n
In [49]: M(i, j, n).args[3] == n
Out[49]: False
Aaron Meurer
On Sep 19, 2010, at 11:55 AM, Aaron S. Meurer wrote:
> Well, now I am having trouble reproducing the problem I had. I was getting:
>
> In [23]: M(i, j, n).args
> ---------------------------------------------------------------------------
> TypeError Traceback (most recent call last)
>
> /Users/aaronmeurer/Documents/Python/sympy/sympy/<ipython console> in
> <module>()
>
> /Users/aaronmeurer/Documents/Python/sympy/sympy/sympy/tensor/indexed.pyc in
> __call__(self, *indices, **kw_args)
> 121
> 122 def __call__(self, *indices, **kw_args):
> --> 123 return IndexedElement(self, *indices, **kw_args)
> 124
> 125 @property
>
> /Users/aaronmeurer/Documents/Python/sympy/sympy/sympy/tensor/indexed.pyc in
> __new__(cls, stem, *args, **kw_args)
> 154
> 155 # FIXME: 2.4 compatibility
>
> --> 156 args = map(_ensure_Idx, args)
> 157 # args = tuple([ a if isinstance(a, Idx) else Idx(a) for a in
> args ])
>
> 158 return Expr.__new__(cls, stem, *args, **kw_args)
>
> /Users/aaronmeurer/Documents/Python/sympy/sympy/sympy/tensor/indexed.pyc in
> _ensure_Idx(arg)
> 135 return arg
> 136 else:
> --> 137 return Idx(arg)
> 138
> 139 class IndexedElement(Expr):
>
> /Users/aaronmeurer/Documents/Python/sympy/sympy/sympy/tensor/indexed.pyc in
> __new__(cls, label, range, **kw_args)
> 235
> 236 if not label.is_integer:
> --> 237 raise TypeError("Idx object requires an integer label")
> 238
> 239 elif isinstance(range, (tuple, list, Tuple)):
>
> TypeError: Idx object requires an integer label
>
> But now I just get
>
> In [21]: M(i, j, n).args
> Out[21]: (M, i, j, n)
>
> By the way, shouldn't it be just (i, j, n), like Function does?
>
> In [20]: f(x, y).args
> Out[20]: (x, y)
>
> But anyway, now that it works, I think I can get somewhere.
>
> Aaron Meurer
>
> On Sep 19, 2010, at 11:20 AM, Aaron S. Meurer wrote:
>
>> So given Chris's tip in issue 2058 (which flew over my head the first time
>> for some reason), I think I might have this figured out, except for one
>> thing. Given M(i, j, n), how to I get the third argument, n? I tried M(i,
>> j, n)[2] and M(i, j, n).args, but neither works.
>>
>> Aaron Meurer
>
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